Using the fact that \(\lbrace \overline{TG}: G \in H^2(\partial \Omega )\rbrace  = H^2(\partial \Omega )^{\perp }\) , we will show that \(f_{\infty } \in H^2(\partial \Omega )\) , cf. [1]}, [2]}. Since
\(\langle f_j^{+}, \overline{T_j G} \rangle _{\partial \Omega _j} = 0\) ,
\({|\langle f_{\infty }, \overline{TG} \rangle |}\\&=& |\langle f_{\infty } - (f_j^{+} \circ \Lambda _j) \omega _j^{1 \over 2}, \overline{TG} \rangle _{\partial \Omega }\\&& \quad + \int _{\partial \Omega } f_j^{+}(\Lambda _j)(w))\omega _j^{1 \over 2}(w)T(w)G(w) - (f_j^{+} (\Lambda _j)(w))T_j(\Lambda _j(w))G(\Lambda _j(w))\omega _j(w) d\sigma (w)|\\&\le & |\langle f_{\infty } - (f_j^{+} \circ \Lambda _j)\omega _j^{1 \over 2}, \overline{TG} \rangle _{\partial \Omega }| + M^{1 \over 2}\Big {(}\int _{\partial \Omega } |T(w)G(w) - T_j(\Lambda _j(w))G(\Lambda _j(w))\omega _j^{1 \over 2}(w)|^2 d\sigma (w)\Big {)}^{1 \over 2}.\) 
